Re: Normal distribution, again

Hey Trefoil2727.

I think the problem is that you have P(Z>z) instead of P(Z<z). Since you are looking at P(-k < X < k) then it means that you are looking at 1 - 2*P(X < k) due to symmetry of the normal distribution which means that 1 - 2*P(Z < z) = 0.8 where z = (k - 5)/2

If you use this you get using R:

> (qnorm(0.1,0,1)*2+5)
[1] 2.436897

Since I used a computer for this, if the answer is based on using tables then it will probably lose precision since the answer of 2.43 is in the tails of the distribution.

Re: Normal distribution, again

Originally Posted by chiro

Hey Trefoil2727.

I think the problem is that you have P(Z>z) instead of P(Z<z). Since you are looking at P(-k < X < k) then it means that you are looking at 1 - 2*P(X < k) due to symmetry of the normal distribution which means that 1 - 2*P(Z < z) = 0.8 where z = (k - 5)/2

If you use this you get using R:

> (qnorm(0.1,0,1)*2+5)
[1] 2.436897

Since I used a computer for this, if the answer is based on using tables then it will probably lose precision since the answer of 2.43 is in the tails of the distribution.

Re: Normal distribution, again

The R package allows you to do a lot of statistical computations very easily.

qnorm(0.1,0,1) returns the value of z where P(Z < z) = 0.1.

You can do the same thing by getting a statistical normal table and finding the value of z where P(Z < z).

The difference is that qnorm uses a computational algorithm where-as a table just gives values (calculating by the same kind of computer algorithm) and you look up those values on paper rather than with a computer.

Re: Normal distribution, again

You seem to have used .08 rather than .8. If you look at the Normal Distribution table here, you will see that "z= .1" corresponds to P(z)= .04, NOT .40. (Since this app is giving P(z> 0), and your problem has -.8< z< .8, you divide by 2 to get 0< z< .4.) P(z< .4) corresponds to z= 1.28, not .1.